Theorem 1. For any two points and a plane, the distance between them is expressed by the formula:
For example, if points and are given, then the distance between them is:
2. Area of a triangle.
Theorem 2.
For any points
not lying on the same straight line, the area of a triangle is expressed by the formula:
For example, let's find the area of the triangle formed by the points , and.
Comment. If the area of a triangle is zero, this means that the points lie on the same line.
3. Division of a segment in a given ratio.
Let an arbitrary segment be given on the plane and let
– any point of this segment other than the end points. The number defined by equality is called attitude, at which the point divides the segment.
The problem of dividing a segment in a given relation is to: for a given relation and given coordinates of points
and find the coordinates of the point.
Theorem 3.
If a point divides a segment
in a relationship
,
then the coordinates of this point are determined by the formulas: 
(1.3), where are the coordinates of the point, and are the coordinates of the point.
Consequence: If is the midpoint of the segment
, where and, then (1.4) (since).
For example. Points and are given. Find the coordinates of a point that is twice closer to than to
Solution: The required point divides the segment
in relation to since 
, Then 
,
, got
Polar coordinates
The most important after the rectangular coordinate system is the polar coordinate system. It consists of a certain point called pole, and the ray emanating from it - polar axis. In addition, the scale unit for measuring the lengths of segments is set. 
Let a polar coordinate system be given and let be an arbitrary point on the plane. Let be the distance from the point
to the point ; – the angle by which the polar axis must be rotated to align with the beam.
Polar coordinates of a point are called numbers. In this case, the number is considered the first coordinate and is called polar radius, the number is the second coordinate and is called polar angle.
Denoted by . The polar radius can have any non-negative value:. It is usually believed that the polar angle varies within the following limits:. However, in some cases it is necessary to determine angles measured from the polar axis clockwise.
The relationship between the polar coordinates of a point and its rectangular coordinates.
We will assume that the origin of the rectangular coordinate system is at the pole, and the positive semi-axis of the abscissa coincides with the polar axis. 
Let – in a rectangular coordinate system and – in a polar coordinate system. Defined – right triangle c. Then(1.5). These formulas express rectangular coordinates in terms of polar ones.
On the other hand, according to the Pythagorean theorem and
(1.6) – these formulas express polar coordinates through rectangular ones.
Note that the formula defines two values of the polar angle, since. From these two angle values, choose the one at which the equalities are satisfied.
For example, let's find the polar coordinates of the point ..or, because I quarter.
Example 1: Find a point symmetrical to a point 
Relative to the bisector of the first coordinate angle.
Solution:
Let's draw through the point A direct l 1, perpendicular to the bisector l first coordinate angle. Let . On a straight line l 1 put aside the segment SA 1 , equal to the segment AC. Right Triangles ASO And A 1 CO equal to each other (on two sides). It follows that | OA| = |O.A. 1 |. Triangles ADO And OEA 1 are also equal to each other (in hypotenuse and acute angle). We conclude that |AD| = |OE| = 4,|OD| = |EA 1 | = 2, i.e. the point has coordinates x = 4, y = -2, those. A 1 (4;-2).
Note that there is a general statement: point A 1, symmetrical to the point relative to the bisector of the first and third coordinate angles, has coordinates, that is .
Example 2: Find the point at which a line passing through the points and , will intersect the axis Oh.
Solution:
Coordinates of the desired point WITH There is ( x; 0). And since the points A,IN And WITH lie on the same straight line, then the condition must be satisfied (x 2 -x 1 )(y 3 -y 1 )-(x 3 -x 1 )(y 2 -y 1 ) = 0 (formula (1.2), area of the triangle ABC equal to zero!), where are the coordinates of the point A, – points IN, – points WITH. We get, i.e., , . Therefore, the point WITH has coordinates ,, i.e..
Example 3: In the polar coordinate system, points are given. Find: A) distance between points and ; b) area of the triangle OM 1 M 2 (ABOUT– pole).
Solution:
a) Let us use formulas (1.1) and (1.5):
that is, .
b) using the formula for the area of a triangle with sides A And b and the angle between them (), we find the area of the triangle OM 1 M 2 . .
Solving problems in mathematics is often accompanied by many difficulties for students. Helping the student cope with these difficulties, as well as teach them to apply their existing theoretical knowledge when solving specific problems in all sections of the course in the subject “Mathematics” is the main purpose of our site.
When starting to solve problems on the topic, students should be able to construct a point on a plane using its coordinates, as well as find the coordinates of a given point.
Calculation of the distance between two points A(x A; y A) and B(x B; y B) taken on a plane is performed using the formula d = √((x A – x B) 2 + (y A – y B) 2), where d is the length of the segment that connects these points on the plane.
If one of the ends of the segment coincides with the origin of coordinates, and the other has coordinates M(x M; y M), then the formula for calculating d will take the form OM = √(x M 2 + y M 2).
1. Calculation of the distance between two points based on the given coordinates of these points
Example 1.
Find the length of the segment that connects points A(2; -5) and B(-4; 3) on the coordinate plane (Fig. 1).
Solution.
The problem statement states: x A = 2; x B = -4; y A = -5 and y B = 3. Find d.
Applying the formula d = √((x A – x B) 2 + (y A – y B) 2), we get:
d = AB = √((2 – (-4)) 2 + (-5 – 3) 2) = 10.
2. Calculation of the coordinates of a point that is equidistant from three given points
Example 2.
Find the coordinates of point O 1, which is equidistant from three points A(7; -1) and B(-2; 2) and C(-1; -5).
Solution.
From the formulation of the problem conditions it follows that O 1 A = O 1 B = O 1 C. Let the desired point O 1 have coordinates (a; b). Using the formula d = √((x A – x B) 2 + (y A – y B) 2) we find:
O 1 A = √((a – 7) 2 + (b + 1) 2);
O 1 B = √((a + 2) 2 + (b – 2) 2);
O 1 C = √((a + 1) 2 + (b + 5) 2).
Let's create a system of two equations:
(√((a – 7) 2 + (b + 1) 2) = √((a + 2) 2 + (b – 2) 2),
(√((a – 7) 2 + (b + 1) 2) = √((a + 1) 2 + (b + 5) 2).
After squaring the left and right sides of the equations, we write:
((a – 7) 2 + (b + 1) 2 = (a + 2) 2 + (b – 2) 2,
((a – 7) 2 + (b + 1) 2 = (a + 1) 2 + (b + 5) 2.
Simplifying, let's write
(-3a + b + 7 = 0,
(-2a – b + 3 = 0.
Having solved the system, we get: a = 2; b = -1.
Point O 1 (2; -1) is equidistant from the three points specified in the condition that do not lie on the same straight line. This point is the center of a circle passing through three given points (Fig. 2).
3. Calculation of the abscissa (ordinate) of a point that lies on the abscissa (ordinate) axis and is at a given distance from a given point
Example 3.
The distance from point B(-5; 6) to point A lying on the Ox axis is 10. Find point A.
Solution.
From the formulation of the problem conditions it follows that the ordinate of point A is equal to zero and AB = 10.
Denoting the abscissa of point A by a, we write A(a; 0).
AB = √((a + 5) 2 + (0 – 6) 2) = √((a + 5) 2 + 36).
We get the equation √((a + 5) 2 + 36) = 10. Simplifying it, we have
a 2 + 10a – 39 = 0.
The roots of this equation are a 1 = -13; and 2 = 3.
We get two points A 1 (-13; 0) and A 2 (3; 0).
Examination:
A 1 B = √((-13 + 5) 2 + (0 – 6) 2) = 10.
A 2 B = √((3 + 5) 2 + (0 – 6) 2) = 10.
Both obtained points are suitable according to the conditions of the problem (Fig. 3).
4. Calculation of the abscissa (ordinate) of a point that lies on the abscissa (ordinate) axis and is at the same distance from two given points
Example 4.
Find a point on the Oy axis that is at the same distance from points A (6, 12) and B (-8, 10).
Solution.
Let the coordinates of the point required by the conditions of the problem, lying on the Oy axis, be O 1 (0; b) (at the point lying on the Oy axis, the abscissa is zero). It follows from the condition that O 1 A = O 1 B.
Using the formula d = √((x A – x B) 2 + (y A – y B) 2) we find:
O 1 A = √((0 – 6) 2 + (b – 12) 2) = √(36 + (b – 12) 2);
O 1 B = √((a + 8) 2 + (b – 10) 2) = √(64 + (b – 10) 2).
We have the equation √(36 + (b – 12) 2) = √(64 + (b – 10) 2) or 36 + (b – 12) 2 = 64 + (b – 10) 2.
After simplification we get: b – 4 = 0, b = 4.
Point O 1 (0; 4) required by the conditions of the problem (Fig. 4).
5. Calculation of the coordinates of a point that is located at the same distance from the coordinate axes and some given point
Example 5.
Find point M located on the coordinate plane at the same distance from the coordinate axes and from point A(-2; 1).
Solution.
The required point M, like point A(-2; 1), is located in the second coordinate angle, since it is equidistant from points A, P 1 and P 2 (Fig. 5). The distances of point M from the coordinate axes are the same, therefore, its coordinates will be (-a; a), where a > 0.
From the conditions of the problem it follows that MA = MR 1 = MR 2, MR 1 = a; MP 2 = |-a|,
those. |-a| = a.
Using the formula d = √((x A – x B) 2 + (y A – y B) 2) we find:
MA = √((-а + 2) 2 + (а – 1) 2).
Let's make an equation:
√((-а + 2) 2 + (а – 1) 2) = а.
After squaring and simplification we have: a 2 – 6a + 5 = 0. Solve the equation, find a 1 = 1; and 2 = 5.
We obtain two points M 1 (-1; 1) and M 2 (-5; 5) that satisfy the conditions of the problem. 
6. Calculation of the coordinates of a point that is located at the same specified distance from the abscissa (ordinate) axis and from the given point
Example 6.
Find a point M such that its distance from the ordinate axis and from point A(8; 6) is equal to 5.
Solution.
From the conditions of the problem it follows that MA = 5 and the abscissa of point M is equal to 5. Let the ordinate of point M be equal to b, then M(5; b) (Fig. 6).
According to the formula d = √((x A – x B) 2 + (y A – y B) 2) we have:
MA = √((5 – 8) 2 + (b – 6) 2).
Let's make an equation:
√((5 – 8) 2 + (b – 6) 2) = 5. Simplifying it, we get: b 2 – 12b + 20 = 0. The roots of this equation are b 1 = 2; b 2 = 10. Consequently, there are two points that satisfy the conditions of the problem: M 1 (5; 2) and M 2 (5; 10).
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Distance from point to point is the length of the segment connecting these points on a given scale. Thus, when it comes to measuring distance, you need to know the scale (unit of length) in which the measurements will be carried out. Therefore, the problem of finding the distance from point to point is usually considered either on a coordinate line or in a rectangular Cartesian coordinate system on a plane or in three-dimensional space. In other words, most often you have to calculate the distance between points using their coordinates.
In this article, we will firstly recall how the distance from point to point on a coordinate line is determined. Next, we obtain formulas for calculating the distance between two points of a plane or space according to given coordinates. In conclusion, we will consider in detail the solutions to typical examples and problems.
Page navigation.
The distance between two points on a coordinate line.
Let's first define the notation. We will denote the distance from point A to point B as .
From this we can conclude that the distance from point A with coordinate to point B with coordinate is equal to the modulus of the difference in coordinates, that is, 
for any location of points on the coordinate line.
Distance from point to point on a plane, formula.
We obtain a formula for calculating the distance between points and given in a rectangular Cartesian coordinate system on a plane.
Depending on the location of points A and B, the following options are possible.
If points A and B coincide, then the distance between them is zero.
If points A and B lie on a straight line perpendicular to the abscissa axis, then the points coincide, and the distance is equal to the distance . In the previous paragraph, we found out that the distance between two points on a coordinate line is equal to the modulus of the difference between their coordinates, therefore, 
. Hence, .
Similarly, if points A and B lie on a straight line perpendicular to the ordinate axis, then the distance from point A to point B is found as .
In this case, triangle ABC is rectangular in construction, and 
And . By Pythagorean theorem we can write down the equality, whence .
Let us summarize all the results obtained: the distance from a point to a point on a plane is found through the coordinates of the points using the formula 
.
The resulting formula for finding the distance between points can be used when points A and B coincide or lie on a straight line perpendicular to one of the coordinate axes. Indeed, if A and B coincide, then . If points A and B lie on a straight line perpendicular to the Ox axis, then. If A and B lie on a straight line perpendicular to the Oy axis, then .
Distance between points in space, formula.
Let us introduce a rectangular coordinate system Oxyz in space. Let's get a formula for finding the distance from a point 
to the point 
.
In general, points A and B do not lie in a plane parallel to one of the coordinate planes. Let us draw through points A and B planes perpendicular to the coordinate axes Ox, Oy and Oz. The intersection points of these planes with the coordinate axes will give us the projections of points A and B onto these axes. We denote the projections 
.
The required distance between points A and B is the diagonal of the rectangular parallelepiped shown in the figure. By construction, the dimensions of this parallelepiped are equal 
And . In a high school geometry course, it was proven that the square of the diagonal of a cuboid is equal to the sum of the squares of its three dimensions, therefore, . Based on the information in the first section of this article, we can write the following equalities, therefore,
where do we get it from formula for finding the distance between points in space .
This formula is also valid if points A and B
- match up;
- belong to one of the coordinate axes or a line parallel to one of the coordinate axes;
- belong to one of the coordinate planes or a plane parallel to one of the coordinate planes.
Finding the distance from point to point, examples and solutions.
So, we have obtained formulas for finding the distance between two points on a coordinate line, plane and three-dimensional space. It's time to look at solutions to typical examples.
The number of problems in which the final step is to find the distance between two points according to their coordinates is truly enormous. Full review Such examples are beyond the scope of this article. Here we will limit ourselves to examples in which the coordinates of two points are known and it is necessary to calculate the distance between them.
Using coordinates, the location of an object on the globe is determined. Coordinates are indicated by latitude and longitude. Latitudes are measured from the equator line on both sides. In the Northern Hemisphere the latitudes are positive, in the Southern Hemisphere they are negative. Longitude is measured from the prime meridian either east or west, respectively, either eastern or western longitude is obtained.According to the generally accepted position, the prime meridian is taken to be the one that passes through the old Greenwich Observatory in Greenwich. Geographic coordinates of the location can be obtained using a GPS navigator. This device receives signals satellite system positioning in the WGS-84 coordinate system, uniform for the whole world.
Navigator models vary in manufacturer, functionality and interface. Currently, built-in GPS navigators are also available in some models cell phones. But any model can record and save the coordinates of a point.
Distance between GPS coordinates
To solve practical and theoretical problems in some industries, it is necessary to be able to determine the distances between points by their coordinates. There are several ways you can do this. Canonical representation form geographical coordinates: degrees, minutes, seconds.For example, you can determine the distance between the following coordinates: point No. 1 - latitude 55°45′07″ N, longitude 37°36′56″ E; point No. 2 - latitude 58°00′02″ N, longitude 102°39′42″ E.
The easiest way is to use a calculator to calculate the length between two points. In the browser search engine, you must set the following search parameters: online - to calculate the distance between two coordinates. In the online calculator, latitude and longitude values are entered into the query fields for the first and second coordinates. When calculating, the online calculator gave the result - 3,800,619 m.
The next method is more labor-intensive, but also more visual. It is necessary to use any available cartographic or navigation program. Programs in which you can create points using coordinates and measure distances between them include the following applications: BaseCamp (a modern analogue of the MapSource program), Google Earth, SAS.Planet.
All of the above programs are available to any network user. For example, to calculate the distance between two coordinates in Google Earth, you need to create two labels indicating the coordinates of the first point and the second point. Then, using the “Ruler” tool, you need to connect the first and second marks with a line, the program will automatically display the measurement result and show the path on the satellite image of the Earth.
In the case of the example given above, the Google Earth program returned the result - the length of the distance between point No. 1 and point No. 2 is 3,817,353 m.
Why there is an error when determining the distance
All calculations of the extent between coordinates are based on the calculation of the arc length. The radius of the Earth is involved in calculating the length of the arc. But since the shape of the Earth is close to an oblate ellipsoid, the radius of the Earth varies at certain points. To calculate the distance between coordinates, the average value of the Earth's radius is taken, which gives an error in the measurement. The greater the distance being measured, the greater the error.There will be a calculator here
Distance between two points on a line
Consider a coordinate line on which 2 points are marked: A A A And B B B. To find the distance between these points, you need to find the length of the segment A B AB A B. This is done using the following formula:
Distance between two points on a lineA B = ∣ a − b ∣ AB=|a-b|A B =∣ a −b∣,
Where a , b a, b a, b- coordinates of these points on a straight line (coordinate line).
Due to the fact that the formula contains a modulus, when solving it, it is not important which coordinate to subtract from which (since the absolute value of this difference is taken).
∣ a − b ∣ = ∣ b − a ∣ |a-b|=|b-a|∣ a −b ∣ =∣ b −a∣
Let's look at an example to better understand the solution to such problems.
Example 1Points are marked on the coordinate line A A A, whose coordinate is equal to 9 9 9 and period B B B with coordinate − 1 -1 − 1 . We need to find the distance between these two points.
Solution
Here a = 9 , b = − 1 a=9, b=-1 a =9, b =− 1
We use the formula and substitute the values:
A B = ∣ a − b ∣ = ∣ 9 − (− 1) ∣ = ∣ 10 ∣ = 10 AB=|a-b|=|9-(-1)|=|10|=10A B =∣ a −b ∣ =∣ 9 − (− 1 ) ∣ = ∣ 1 0 ∣ = 1 0
Answer
Distance between two points on a plane
Consider two points given on a plane. From each point marked on the plane, you need to lower two perpendiculars: To the axis O X OX O X and on the axle O Y OY OY. Then the triangle is considered A B C ABC A B C. Since it is rectangular ( B C BC B C perpendicular A C AC A C), then find the segment A B AB A B, which is also the distance between points, can be done using the Pythagorean theorem. We have:
A B 2 = A C 2 + B C 2 AB^2=AC^2+BC^2A B 2 = A C 2 + B C 2
But, based on the fact that the length A C AC A C equal to x B − x A x_B-x_A x B − x A , and the length B C BC B C equal to y B − y A y_B-y_A y B − y A , this formula can be rewritten as follows:
Distance between two points on a planeA B = (x B − x A) 2 + (y B − y A) 2 AB=\sqrt((x_B-x_A)^2+(y_B-y_A)^2)A B =(x B − x A ) 2 + (y B − y A ) 2 ,
Where x A , y A x_A, y_A x A , y A And x B , y B x_B, y_B x B , y B - coordinates of points A A A And B B B respectively.
Example 2It is necessary to find the distance between points C C C And F F F, if the coordinates of the first (8 ; − 1) (8;-1) (8 ; − 1 ) , and second - (4 ; 2) (4;2) (4 ; 2 ) .
Solution
X C = 8 x_C=8 x C
=
8
y C = − 1 y_C=-1 y C
=
−
1
x F = 4 x_F=4 x F
=
4
y F = 2 y_F=2 y F
=
2
C F = (x F − x C) 2 + (y F − y C) 2 = (4 − 8) 2 + (2 − (− 1)) 2 = 16 + 9 = 25 = 5 CF=\sqrt(( x_F-x_C)^2+(y_F-y_C)^2)=\sqrt((4-8)^2+(2-(-1))^2)=\sqrt(16+9)=\sqrt( 25)=5C F =(x F − x C ) 2 + (y F − y C ) 2 = (4 − 8 ) 2 + (2 − (− 1 ) ) 2 = 1 6 + 9 = 2 5 = 5
Answer
Distance between two points in space
Finding the distance between two points in this case is similar to the previous one, except that the coordinates of the point in space are specified by three numbers; accordingly, the coordinate of the applicate axis must also be added to the formula. The formula will look like this:
Distance between two points in spaceA B = (x B − x A) 2 + (y B − y A) 2 + (z B − z A) 2 AB=\sqrt((x_B-x_A)^2+(y_B-y_A)^2+( z_B-z_A)^2)A B =(x B − x A ) 2 + (y B − y A ) 2 + (z B − zA ) 2
Example 3Find the length of the segment FK FK
Solution
F = (− 1 ; − 1 ; 8) F=(-1;-1;8)
F K = (x K − x F) 2 + (y K − y F) 2 + (z K − z F) 2 = (− 3 − (− 1)) 2 + (6 − (− 1)) 2 + (0 − 8) 2 = 117 ≈ 10.8 FK=\sqrt((x_K-x_F)^2+(y_K-y_F)^2+(z_K-z_F)^2)=\sqrt((-3-(-1 ))^2+(6-(-1))^2+(0-8)^2)=\sqrt(117)\approx10.8
According to the conditions of the problem, we need to round the answer to a whole number.
